ECEN 301 Discussion #20 – Exam 2 Review 1
Date Day ClassNo.
Title Chapters HWDue date
LabDue date
Exam
10 Nov Mon 20 Exam Review
LAB 7
EXAM 2
11 Nov Tue
12 Nov Wed 21 Boolean Algebra 13.2 – 13.3
13 Nov Thu
14 Nov Fri Recitation
15 Nov Sat
16 Nov Sun
17 Nov Mon 22 Combinational Logic 13.3 – 13.5
LAB 1018 Nov Tue
Schedule…
ECEN 301 Discussion #20 – Exam 2 Review 2
Ask
Alma 5:26 26 And now behold, I say unto you, my
brethren, if ye have experienced a change of heart, and if ye have felt to sing the song of redeeming love, I would ask, can ye feel so now?
ECEN 301 Discussion #20 – Exam 2 Review 3
Lecture 20 – Exam 2 Review
Chapters 4 – 6, 8
ECEN 301 Discussion #20 – Exam 2 Review 4
Exam 2 12 – 16 November (Monday – Friday) Chapters 4 – 6 and 8 15 questions
12 multiple choice (answer on bubble sheet!)• 1 point each
3 long answer (show your work!)• 4 or 5 points each
Closed book! One 3x5 card allowed
Calculators allowed No time limit Study lecture slides and homework
ECEN 301 Discussion #20 – Exam 2 Review 5
Exam 2 Review…Overview
1. Capacitors and Inductors2. Measuring Signal Strength3. Phasors4. Impedance5. AC RLC Circuits6. AC Equivalent Circuits7. DC Transient Response8. Frequency Response9. Basic Filters10. Op-Amps
ECEN 301 Discussion #20 – Exam 2 Review 6
Capacitors & Inductors
Inductors Capacitors
Passive sign convention
Voltage
Current
Power
)()(1
)( 00
tidvL
tit
t L
dt
tdiLtv
)()(
+ L –
i
+ C –
i
)()(1
)( 00
tvdiC
tvt
t C
dt
tdvCti
)()(
dt
tditLitPL
)()()(
dt
tdvtCvtPC
)()()(
ECEN 301 Discussion #20 – Exam 2 Review 7
Capacitors & Inductors
Inductors Capacitors
Energy
An instantaneous change is not permitted in:
Current Voltage
Will permit an instantaneous change in:
Voltage Current
With DC source element acts as a:
Short Circuit Open Circuit
2)(2
1)( tLitWL 2)(
2
1)( tCvtWC
ECEN 301 Discussion #20 – Exam 2 Review 8
Capacitors & Inductors
1. What is the difference between the voltage and current behaviour of capacitors and inductors?
ECEN 301 Discussion #20 – Exam 2 Review 9
Capacitors & Inductors
0.0
1.0
2.0
3.0
4.0
5.0
0.00 2.00 4.00 6.00
0.0
0.1
0.2
0.3
0.4
0.5
0.00 2.00 4.00 6.00
Capacitor voltage vC(t)Inductor current iL(t)
NB: neither can change instantaneously
Capacitor current iC(t)Inductor voltage vL(t)
NB: both can change instantaneously
1. What is the difference between the voltage and current behaviour of capacitors and inductors?
ECEN 301 Discussion #20 – Exam 2 Review 10
Capacitors & Inductors
2. find the voltage v(t) for a capacitor C = 0.5F with the current as shown and v(0) = 0
0.0
0.2
0.4
0.6
0.8
1.0
1.2
0.0 1.0 2.0 3.0
time (s)
curr
ent
(A)
ECEN 301 Discussion #20 – Exam 2 Review 11
Capacitors & Inductors
0.0
0.2
0.4
0.6
0.8
1.0
1.2
0.0 1.0 2.0 3.0
time (s)
curr
ent
(A)
t
t
tt
ti
20
211
10
00
:intervals 4in i(t)current
)0(1
)(0
vidC
tvt
2. find the voltage v(t) for a capacitor C = 0.5F with the current as shown and v(0) = 0
ECEN 301 Discussion #20 – Exam 2 Review 12
Capacitors & Inductors
0.0
0.2
0.4
0.6
0.8
1.0
1.2
0.0 1.0 2.0 3.0
time (s)
curr
ent
(A)
tv
tvd
td
tv
t
t
2)2(0
21)1()1(2
1002
00
:intervals 4in v(t)voltage
1
0
)0(1
)(0
vidC
tvt
2. find the voltage v(t) for a capacitor C = 0.5F with the current as shown and v(0) = 0
ECEN 301 Discussion #20 – Exam 2 Review 13
Capacitors & Inductors
0.0
0.2
0.4
0.6
0.8
1.0
1.2
0.0 1.0 2.0 3.0
time (s)
curr
ent
(A)
t
tt
tt
tv
23
2112
10
00
:intervals 4in v(t)voltage
2
2. find the voltage v(t) for a capacitor C = 0.5F with the current as shown and v(0) = 0
ECEN 301 Discussion #20 – Exam 2 Review 14
Capacitors & Inductors
0.0
0.5
1.0
1.5
2.0
2.5
3.0
3.5
0.0 1.0 2.0 3.0 4.0
time (s)
vo
lta
ge
(V
)
0.0
0.2
0.4
0.6
0.8
1.0
1.2
0.0 1.0 2.0 3.0
time (s)
curr
ent
(A)
t
tt
tt
tv
23
2112
10
00
:intervals 4in v(t)voltage
2
NB: The final value of the capacitor voltage after the current source has stopped charging the capacitor depends on two things:1. The initial capacitor voltage2. The history of the capacitor current
2. find the voltage v(t) for a capacitor C = 0.5F with the current as shown and v(0) = 0
ECEN 301 Discussion #20 – Exam 2 Review 15
Measuring Signal Strength
3. Compute the rms value of the sinusoidal current i(t) = I cos(ωt)
ECEN 301 Discussion #20 – Exam 2 Review 16
Measuring Signal Strength
2
02
1
)2cos(222
1
)2cos(2
1
2
1
2
)(cos2
)(1
2
/2
0
22
/2
0
2
/2
0
22
0
2
I
I
dI
I
dI
dI
diT
iT
rms
Integrating a sinusoidal waveform over 2 periods equals zero
2
1)2cos()(cos2
tt
3. Compute the rms value of the sinusoidal current i(t) = I cos(ωt)
ECEN 301 Discussion #20 – Exam 2 Review 17
Measuring Signal Strength
2
02
1
)2cos(222
1
)2cos(2
1
2
1
2
)2(cos2
)(1
2
/2
0
22
/2
0
2
/2
0
22
0
2
I
I
dI
I
dI
dI
diT
iT
rms
The RMS value of any sinusoid signal is always equal to 0.707 times the peak value (regardless of phase or frequency)
3. Compute the rms value of the sinusoidal current i(t) = I cos(ωt)
ECEN 301 Discussion #20 – Exam 2 Review 18
Phasors4. compute the phasor voltage for the equivalent voltage vs(t)
v1(t) = 15cos(377t+π/4)
v2(t) = 15cos(377t+π/12)
v1(t)+–~
v2(t)+–~
vs(t)+–~
ECEN 301 Discussion #20 – Exam 2 Review 19
Phasors
v1(t)+–~
v2(t)+–~
vs(t)+–~
1. Write voltages in phasor notation
V
ejV
V
ejV
j
j
1215
15)(
415
15)(
12/2
4/1
4. compute the phasor voltage for the equivalent voltage vs(t)v1(t) = 15cos(377t+π/4)
v2(t) = 15cos(377t+π/12)
ECEN 301 Discussion #20 – Exam 2 Review 20
Phasors
Vj
jjV
VjV
88.349.14
12sin15
12cos15)(
:rrectangula Convert to12
15)(
2
2
v1(t)+–~
v2(t)+–~
vs(t)+–~
1. Write voltages in phasor notation2. Convert phasor voltages from polar to
rectangular form (see Appendix A)
Vj
jjV
VjV
61.1061.10
4sin15
4cos15)(
:rrectangula Convert to4
15)(
1
1
4. compute the phasor voltage for the equivalent voltage vs(t)v1(t) = 15cos(377t+π/4)
v2(t) = 15cos(377t+π/12)
ECEN 301 Discussion #20 – Exam 2 Review 21
Phasors
v1(t)+–~
v2(t)+–~
vs(t)+–~
1. Write voltages in phasor notation2. Convert phasor voltages from polar to
rectangular form (see Appendix A)3. Combine voltages
49.1410.25
)()()( 21
j
jVjVjVS
4. compute the phasor voltage for the equivalent voltage vs(t)v1(t) = 15cos(377t+π/4)
v2(t) = 15cos(377t+π/12)
ECEN 301 Discussion #20 – Exam 2 Review 22
Phasors
v1(t)+–~
v2(t)+–~
vs(t)+–~
1. Write voltages in phasor notation2. Convert phasor voltages from polar to
rectangular form (see Appendix A)3. Combine voltages4. Convert rectangular back to polar
698.28)(
6
10.25
49.14tan
98.28
(14.49)(25.10) r
:polar toConvert
49.1410.25)(
1
22
jV
jjV
S
S
4. compute the phasor voltage for the equivalent voltage vs(t)v1(t) = 15cos(377t+π/4)
v2(t) = 15cos(377t+π/12)
ECEN 301 Discussion #20 – Exam 2 Review 23
Phasors
v1(t)+–~
v2(t)+–~
vs(t)+–~
1. Write voltages in phasor notation2. Convert phasor voltages from polar to
rectangular form (see Appendix A)3. Combine voltages4. Convert rectangular back to polar5. Convert from phasor to time domain
6377cos98.28)(
698.28)(
ttv
jV
S
S
Bring ωt back
NB: the answer is NOT simply the addition of the amplitudes of v1(t) and v2(t) (i.e. 15 + 15), and the addition of their phases (i.e. π/4 + π/12)
4. compute the phasor voltage for the equivalent voltage vs(t)v1(t) = 15cos(377t+π/4)
v2(t) = 15cos(377t+π/12)
ECEN 301 Discussion #20 – Exam 2 Review 24
Phasors
v1(t)+–~
v2(t)+–~
vs(t)+–~
6377cos98.28)(
698.28)(
ttv
jV
S
S
Re
Im
14.49 π/6
25.10
Vs(jω)
4. compute the phasor voltage for the equivalent voltage vs(t)v1(t) = 15cos(377t+π/4)
v2(t) = 15cos(377t+π/12)
ECEN 301 Discussion #20 – Exam 2 Review 25
ImpedanceImpedance: complex resistance (has no physical significance)
will allow us to use network analysis methods such as node voltage, mesh current, etc.
Capacitors and inductors act as frequency-dependent resistors
vs(t)+–~ R
+vR(t)
–
i(t)
vs(t)+–~ C
+vC(t)
–
i(t)
vs(t)+–~ L
+vL(t)
–
i(t)
Vs(jω) +–~
+VZ(jω)
–
I(jω)
Z
ECEN 301 Discussion #20 – Exam 2 Review 26
Impedance
LjjZL )(RjZR )(
+L–
+C–
Re
Im
-π/2
π/2R
-1/ωC
ωL
ZR
ZC
ZL
Phasor domain
+R–
CjjZC 1
)(
Vs(jω) +–~
+VZ(jω)
–
I(jω)
Z
Impedance of resistors, inductors, and capacitors
ECEN 301 Discussion #20 – Exam 2 Review 27
Impedance
5. find the equivalent impedance (ZEQ)
ω = 104 rads/s, C = 10uF, R1 = 100Ω, R2 = 50Ω, L = 10mH
ZEQ
R2 C
R1
L
ECEN 301 Discussion #20 – Exam 2 Review 28
Impedance
ZEQ
R2 C
R1
L
)37.1(81.9
62.992.1
51
50
)50)(1010)(10(1
50
1
)/1(
)/1(
||
64
2
2
2
2
21
j
j
j
CRj
R
CjR
CjR
ZZZ CREQ
5. find the equivalent impedance (ZEQ)
ω = 104 rads/s, C = 10uF, R1 = 100Ω, R2 = 50Ω, L = 10mH
ECEN 301 Discussion #20 – Exam 2 Review 29
Impedance
723.02.136
38.9092.101
62.992.1)10)(10(100
)37.1(81.924
1
11
j
jj
LjR
ZZZZ EQLREQ
ZEQ
R1
L
ZEQ1
)37.1(81.91EQZNB: at this frequency (ω) the circuit has an inductive impedance (reactance or phase is positive)
5. find the equivalent impedance (ZEQ)
ω = 104 rads/s, C = 10uF, R1 = 100Ω, R2 = 50Ω, L = 10mH
ECEN 301 Discussion #20 – Exam 2 Review 30
AC RLC CircuitsAC Circuit Analysis
1. Identify the AC sources and note the excitation frequency (ω)
2. Convert all sources to the phasor domain3. Represent each circuit element by its impedance4. Solve the resulting phasor circuit using network
analysis methods5. Convert from the phasor domain back to the time
domain
ECEN 301 Discussion #20 – Exam 2 Review 31
AC RLC Circuits
6. find ia(t) and ib(t)vs(t) = 15cos(1500t)V, R1 = 100Ω, R2 = 75Ω, L = 0.5H, C = 1uF
R1
vs(t)+–~
R2
L
C
ia(t) ib(t)
ECEN 301 Discussion #20 – Exam 2 Review 32
AC RLC Circuits
6. find ia(t) and ib(t)vs(t) = 15cos(1500t)V, R1 = 100Ω, R2 = 75Ω, L = 0.5H, C = 1uF
R1
vs(t)+–~
R2
L
C
ia(t) ib(t)
1. Note frequencies of AC sources
Only one AC source - ω = 1500 rad/s
ECEN 301 Discussion #20 – Exam 2 Review 33
AC RLC Circuits
6. find ia(t) and ib(t)vs(t) = 15cos(1500t)V, R1 = 100Ω, R2 = 75Ω, L = 0.5H, C = 1uF
R1
vs(t)+–~
R2
L
C
ia(t) ib(t)
1. Note frequencies of AC sources2. Convert to phasor domain
ZR1
Vs(jω)+–~
ZR2
Ia(jω)
ZL
ZC
Ib(jω)
ECEN 301 Discussion #20 – Exam 2 Review 34
AC RLC Circuits
6. find ia(t) and ib(t)vs(t) = 15cos(1500t)V, R1 = 100Ω, R2 = 75Ω, L = 0.5H, C = 1uF
ZR1
Vs(jω)+–~
ZR2
Ia(jω)
ZL
ZC
Ib(jω)
1. Note frequencies of AC sources2. Convert to phasor domain3. Represent each element by its impedance
10011 RZR
015015)(
)cos(15)(j
s
s
ejV
ttv
7522 RZR
750
)5.0)(1500(
j
j
LjZL
667
)10)(1500(/1
/16
j
j
CjZC
ECEN 301 Discussion #20 – Exam 2 Review 35
AC RLC Circuits
6. find ia(t) and ib(t)vs(t) = 15cos(1500t)V, R1 = 100Ω, R2 = 75Ω, L = 0.5H, C = 1uF
0)(
0)(
0)()()(
:bat KVL
2
2
2
RLCbCa
RbLbCba
RLC
ZZZIZI
ZIZIZII
jVjVjV
+ZR1–
Vs(jω)+–~
+ZR2
–Ia(jω)
+ZL–
+ZC
– Ib(jω)
667
750
75
100
2
1
jZ
jZ
Z
Z
C
L
R
R
4. Solve using network analysis• Mesh current
sCbCRa
sCbaRa
CRs
VZIZZI
VZIIZI
jVjVjV
)(
)(
0)()()(
:aat KVL
1
1
1
ECEN 301 Discussion #20 – Exam 2 Review 36
AC RLC Circuits
6. find ia(t) and ib(t)vs(t) = 15cos(1500t)V, R1 = 100Ω, R2 = 75Ω, L = 0.5H, C = 1uF
0)8375()667( jIjI ba
+ZR1–
Vs(jω)+–~
+ZR2
–Ia(jω)
+ZL–
+ZC
– Ib(jω)
AI
AI
49.1019.0
917.00032.0
2
1
4. Solve using network analysis• Mesh current
15)667()667100( jIjI ba
ECEN 301 Discussion #20 – Exam 2 Review 37
AC RLC Circuits
6. find ia(t) and ib(t)vs(t) = 15cos(1500t)V, R1 = 100Ω, R2 = 75Ω, L = 0.5H, C = 1uF
+ZR1–
Vs(jω)+–~
+ZR2
–Ia(jω)
+ZL–
+ZC
– Ib(jω) mAtti
AI
)917.01500cos(2.3)(
917.00032.0
1
1
5. Convert to Time domain
mAtti
AI
)49.11500cos(19)(
49.1019.0
2
2
ECEN 301 Discussion #20 – Exam 2 Review 38
AC Equivalent CircuitsThévenin and Norton equivalent circuits apply in AC analysis
Equivalent voltage/current will be complex and frequency dependent
Load
+V–
I
Source
VT(jω)+–
ZT
Load
+
V
–
I
IN(jω)
ZN Load
+
V
–
I
Norton EquivalentThévenin Equivalent
ECEN 301 Discussion #20 – Exam 2 Review 39
AC Equivalent CircuitsComputation of Thévenin and Norton Impedances:
1. Remove the load (open circuit at load terminal)2. Zero all independent sources
Voltage sources short circuit (v = 0) Current sources open circuit (i = 0)
3. Compute equivalent impedance across load terminals (with load removed)
NB: same procedure as equivalent resistance
ZL
Z1
Vs(jω)
+–
Z3
Z2
Z4
a
b
Z1 Z3
Z2
Z4
a
b
ZT
ECEN 301 Discussion #20 – Exam 2 Review 40
AC Equivalent CircuitsComputing Thévenin voltage:
1. Remove the load (open circuit at load terminals)2. Define the open-circuit voltage (Voc) across the load terminals3. Chose a network analysis method to find Voc
node, mesh, superposition, etc.
4. Thévenin voltage VT = Voc
Z1
Vs(jω)
+–
Z3
Z2
Z4
a
b
Z1
Vs(jω)
+–
Z3
Z2
Z4
a
b
+VT
–
NB: same procedure as equivalent resistance
ECEN 301 Discussion #20 – Exam 2 Review 41
AC Equivalent CircuitsComputing Norton current:
1. Replace the load with a short circuit2. Define the short-circuit current (Isc) across the load terminals3. Chose a network analysis method to find Isc
node, mesh, superposition, etc.
4. Norton current IN = Isc
Z1
Vs(jω)
+–
Z3
Z2
Z4
a
b
Z1
Vs(jω)
+–
Z3
Z2
Z4
a
b
IN
NB: same procedure as equivalent resistance
ECEN 301 Discussion #20 – Exam 2 Review 42
AC Equivalent Circuits
7. find the Thévenin equivalentω = 103 Hz, Rs = 50Ω, RL = 50Ω, L = 10mH, C = 0.1uF
Rs
vs(t)+–~ RL
L
C +vL
–
ECEN 301 Discussion #20 – Exam 2 Review 43
AC Equivalent Circuits
7. find the Thévenin equivalentω = 103 Hz, Rs = 50Ω, RL = 50Ω, L = 10mH, C = 0.1uF
Rs
vs(t)+–~ RL
L
C +vL
–
1. Note frequencies of AC sources
Only one AC source - ω = 103 rad/s
ECEN 301 Discussion #20 – Exam 2 Review 44
AC Equivalent Circuits
7. find the Thévenin equivalentω = 103 Hz, Rs = 50Ω, RL = 50Ω, L = 10mH, C = 0.1uF
Rs
vs(t)+–~ RL
L
C +vL
–
1. Note frequencies of AC sources2. Convert to phasor domain
Zs
ZLD
+–~
ZL
ZC
Vs(jω)
ECEN 301 Discussion #20 – Exam 2 Review 45
AC Equivalent Circuits
7. find the Thévenin equivalentω = 103 Hz, Rs = 50Ω, RL = 50Ω, L = 10mH, C = 0.1uF
1. Note frequencies of AC sources2. Convert to phasor domain3. Find ZT
• Remove load & zero sources
Zs
ZL
ZC
9182.033.82
414.65501
)/1()(
)/1)((
||
2
jLC
LjR
CjLj
CjLjR
ZZZZ
S
S
LCST
ECEN 301 Discussion #20 – Exam 2 Review 46
AC Equivalent Circuits
7. find the Thévenin equivalentω = 103 Hz, Rs = 50Ω, RL = 50Ω, L = 10mH, C = 0.1uF
9182.033.82 TZ
1. Note frequencies of AC sources2. Convert to phasor domain3. Find ZT
• Remove load & zero sources4. Find VT(jω)
• Remove load
Zs
+–~
ZL
ZC
Vs(jω)+
VT(jω)–
NB: Since no current flows in the circuit once the load is removed:
ST VV
ECEN 301 Discussion #20 – Exam 2 Review 47
AC Equivalent Circuits
7. find the Thévenin equivalentω = 103 Hz, Rs = 50Ω, RL = 50Ω, L = 10mH, C = 0.1uF
9182.033.82 TZST VV
Zs
ZLD
+–~
ZL
ZC
Vs(jω)
ZT
+–~
VT(jω)
ZLD
ECEN 301 Discussion #20 – Exam 2 Review 48
DC Transient Response
Transient response of a circuit consists of 3 parts:1. Steady-state response prior to the switching on/off of a
DC source2. Transient response – the circuit adjusts to the DC source3. Steady-state response following the transient response
R1
R2 Cvs+–
t = 0
DC Source
Switch Energy element
ECEN 301 Discussion #20 – Exam 2 Review 49
Initial condition x(0): DC steady state before a switch is first activated x(0–): right before the switch is closed x(0+): right after the switch is closed
Final condition x(∞): DC steady state a long time after a switch is activated
R1
R2
Cvs
+–
t = 0
R3
R1
R2
Cvs
+–
t → ∞
R3
Initial condition Final condition
DC Transient Response – DC Steady-State
ECEN 301 Discussion #20 – Exam 2 Review 50
Remember – capacitor voltages and inductor currents cannot change instantaneouslyCapacitor voltages and inductor currents don’t change right
before closing and right after closing a switch
)0()0(
)0()0(
LL
CC
ii
vv
DC Transient Response – DC Steady-State
ECEN 301 Discussion #20 – Exam 2 Review 51
DC Transient Response – DC Steady-State
8. find the initial and final current conditions at the inductoris = 10mA
is
t = 0
RLiL
ECEN 301 Discussion #20 – Exam 2 Review 52
8. find the initial and final current conditions at the inductoris = 10mA
is
t = 0
RLiL
1. Initial conditions – assume the current across the inductor is in steady-state.
isiL
NB: in DC steady state inductors act like short circuits, thus no current flows through R
DC Transient Response – DC Steady-State
ECEN 301 Discussion #20 – Exam 2 Review 53
8. find the initial and final current conditions at the inductoris = 10mA 1. Initial conditions – assume the current
across the inductor is in steady-state.
mA
ii sL
10
)0(
is
iL
DC Transient Response – DC Steady-State
ECEN 301 Discussion #20 – Exam 2 Review 54
8. find the initial and final current conditions at the inductoris = 10mA
is
t = 0
RLiL
1. Initial conditions – assume the current across the inductor is in steady-state.
2. Throw the switch
NB: inductor current cannot change instantaneously
RLiL
DC Transient Response – DC Steady-State
ECEN 301 Discussion #20 – Exam 2 Review 55
8. find the initial and final current conditions at the inductoris = 10mA
is
t = 0
RLiL
1. Initial conditions – assume the current across the inductor is in steady-state.
2. Throw the switch
NB: inductor current cannot change instantaneously
–R+
LiL
NB: polarity of R
mA
ii LL
10
)0()0(
DC Transient Response – DC Steady-State
ECEN 301 Discussion #20 – Exam 2 Review 56
8. find the initial and final current conditions at the inductoris = 10mA
is
t = 0
RLiL
1. Initial conditions – assume the current across the inductor is in steady-state.
2. Throw the switch3. Final conditions
NB: in DC steady state inductors act like short circuits
AiL 0)(
DC Transient Response – DC Steady-State
ECEN 301 Discussion #20 – Exam 2 Review 57
DC Transient ResponseSolving 1st order transient response:
1. Solve the DC steady-state circuit: Initial condition x(0–): before switching (on/off) Final condition x(∞): After any transients have died out (t → ∞)
2. Identify x(0+): the circuit initial conditions Capacitors: vC(0+) = vC(0–) Inductors: iL(0+) = iL(0–)
3. Write a differential equation for the circuit at time t = 0+
Reduce the circuit to its Thévenin or Norton equivalent The energy storage element (capacitor or inductor) is the load
The differential equation will be either in terms of vC(t) or iL(t) Reduce this equation to standard form
4. Solve for the time constant Capacitive circuits: τ = RTC Inductive circuits: τ = L/RT
5. Write the complete response in the form: x(t) = x(∞) + [x(0) - x(∞)]e-t/τ
ECEN 301 Discussion #20 – Exam 2 Review 58
DC Transient Response9. find vc(t) for all t
vs = 12V, vC(0–) = 5V, R = 1000Ω, C = 470uF
R
Cvs+–
t = 0
i(t)
+vC(t)
–
ECEN 301 Discussion #20 – Exam 2 Review 59
9. find vc(t) for all tvs = 12V, vC(0–) = 5V, R = 1000Ω, C = 470uF
V
vv SC
12
)(
R
Cvs+–
t = 0
i(t)
+vC(t)
–
1. DC steady-statea) Initial condition: vC(0)b) Final condition: vC(∞)
Vvtv CC 5)0()0(
NB: as t → ∞ the capacitor acts like an open circuit thus vC(∞) = vS
DC Transient Response
ECEN 301 Discussion #20 – Exam 2 Review 60
9. find vc(t) for all tvs = 12V, vC(0–) = 5V, R = 1000Ω, C = 470uF
R
Cvs+–
t = 0
i(t)
+vC(t)
–
2. Circuit initial conditions: vC(0+)
V
vv CC
5
)0()0(
DC Transient Response
ECEN 301 Discussion #20 – Exam 2 Review 61
9. find vc(t) for all tvs = 12V, vC(0–) = 5V, R = 1000Ω, C = 470uF
R
Cvs+–
t = 0
i(t)
+vC(t)
–
3. Write differential equation (already in Thévenin equivalent) at t = 0
SCC
SCC
CRS
vtvdt
tdvRC
vtvRti
tvtvv
)()(
)()(
0)()(
:KVL
DC Transient Response
ECEN 301 Discussion #20 – Exam 2 Review 62
9. find vc(t) for all tvs = 12V, vC(0–) = 5V, R = 1000Ω, C = 470uF
47.0
)10470)(1000( 6
RC
R
Cvs+–
t = 0
i(t)
+vC(t)
–
4. Find the time constant τ
1SK
12 SvF
DC Transient Response
ECEN 301 Discussion #20 – Exam 2 Review 63
9. find vc(t) for all tvs = 12V, vC(0–) = 5V, R = 1000Ω, C = 470uF
R
Cvs+–
t = 0
i(t)
+vC(t)
–
5. Write the complete responsex(t) = x(∞) + [x(0) - x(∞)]e-t/τ
47.0/
47.0/
/
712
)125(12
)()0()()(
t
t
tCCCC
e
e
evvvtv
DC Transient Response
ECEN 301 Discussion #20 – Exam 2 Review 64
Frequency Response
Frequency Response H(jω): a measure of how the voltage/current/impedance of a load responds to the voltage/current of a source
)()(
)(jVjV
jHS
LV
)()(
)(jIjI
jHS
LI
)()(
)(jIjV
jHS
LZ
ECEN 301 Discussion #20 – Exam 2 Review 65
Frequency Response
10. compute the frequency response HV(jω)
R1 = 1kΩ, RL = 10kΩ, C = 10uF
R1
Cvs(t)+–
+RL
–
ECEN 301 Discussion #20 – Exam 2 Review 66
Frequency Response
10. compute the frequency response HV(jω)
R1 = 1kΩ, RL = 10kΩ, C = 10uF
R1
Cvs(t)+–
+RL
–
1. Note frequencies of AC sources
Only one AC source so frequency response HV(jω) will be the function of a single frequency
ECEN 301 Discussion #20 – Exam 2 Review 67
Frequency Response
10. compute the frequency response HV(jω)
R1 = 1kΩ, RL = 10kΩ, C = 10uF
R1
Cvs(t)+–
+RL
–
1. Note frequencies of AC sources2. Convert to phasor domain
Z1 = R1
ZLD=RL
ZC=1/jωC
Vs+–~
ECEN 301 Discussion #20 – Exam 2 Review 68
Frequency Response
10. compute the frequency response HV(jω)
R1 = 1kΩ, RL = 10kΩ, C = 10uF1. Note frequencies of AC sources2. Convert to phasor domain3. Solve using network analysis
• Thévenin equivalent
Z1 = R1
ZC=1/jωC
CT ZZZ ||1
ECEN 301 Discussion #20 – Exam 2 Review 69
Frequency Response
10. compute the frequency response HV(jω)
R1 = 1kΩ, RL = 10kΩ, C = 10uF
C
CST ZZ
ZjVjV
1
)()(
1. Note frequencies of AC sources2. Convert to phasor domain3. Solve using network analysis
• Thévenin equivalent
Z1 = R1
+VT
–
ZC=1/jωC
Vs+–~
CT ZZZ ||1
ECEN 301 Discussion #20 – Exam 2 Review 70
Frequency Response
10. compute the frequency response HV(jω)
R1 = 1kΩ, RL = 10kΩ, C = 10uF
C
CST ZZ
ZjVjV
1
)()(
1. Note frequencies of AC sources2. Convert to phasor domain3. Solve using network analysis
• Thévenin equivalent4. Find an expression for the load voltage
CT ZZZ ||1
ZT
ZLDVT
+–~
+VL
–
LDC
LD
C
CS
LDT
LDTL
ZZZ
Z
ZZ
ZjV
ZZ
ZjVjV
||)(
)()(
11
ECEN 301 Discussion #20 – Exam 2 Review 71
Frequency Response
10. compute the frequency response HV(jω)
R1 = 1kΩ, RL = 10kΩ, C = 10uF5. Find an expression for the frequency
response ZT
ZLDVT
+–~
+VL
–
LDC
LD
C
C
S
LV
ZZZ
Z
ZZ
Z
jV
jVjH
||
)(
)()(
11
ECEN 301 Discussion #20 – Exam 2 Review 72
Frequency Response
10. compute the frequency response HV(jω)
R1 = 1kΩ, RL = 10kΩ, C = 10uF5. Find an expression for the frequency
response ZT
ZLDVT
+–~
+VL
–
110arctan
110
100
110
100
)10/(10)10/(11010
)10/(10
||)(
22
53534
54
11
11
j
jj
j
ZZZZZ
ZZ
ZZZ
Z
ZZ
ZjH
CCLD
CLD
LDC
LD
C
CV
ECEN 301 Discussion #20 – Exam 2 Review 73
Frequency Response
10. compute the frequency response HV(jω)
R1 = 1kΩ, RL = 10kΩ, C = 10uF
2010.0)(
jHV
5. Find an expression for the frequency response• Look at response for low frequencies (ω = 10)
and high frequencies (ω = 10000)
ZT
ZLDVT
+–~
+VL
–
110arctan
110
100)(
22
jHV
0907.0905.0)( jHV
ω = 10ω = 10000
ECEN 301 Discussion #20 – Exam 2 Review 74
Basic Filters
Electric circuit filter: attenuates (reduces) or eliminates signals at unwanted frequencies
0.00. 00 10. 00
.
Low-pass
0.00. 00 10. 00
.
High-pass
0.00.00 10.00
.
Band-pass
0.00. 00 10. 00
.
Band-stop
ω ω
ω ω
ECEN 301 Discussion #20 – Exam 2 Review 75
Basic Filters – Resonant Frequency
Resonant Frequency (ωn): the frequency at which capacitive impedance and inductive impedance are equal and opposite (in 2nd order filters)
LCn
1
C+
vi(t)–
+vo(t)
–
L
C+
vi(t)–
+vo(t)
–L
Impedances in series Impedances in parallel
ECEN 301 Discussion #20 – Exam 2 Review 76
Basic Filters – Resonant Frequency
Resonant Frequency (ωn): the frequency at which capacitive impedance and inductive impedance are equal and opposite (in 2nd order filters)
0)( jVo
C
LCjZL
+Vi(jω)
–
+Vo(jω)
–
ZEQ=0C
LCjZC
0
CLEQ ZZZ
Impedances in series
ECEN 301 Discussion #20 – Exam 2 Review 77
Basic Filters – Resonant Frequency
Resonant Frequency (ωn): the frequency at which capacitive impedance and inductive impedance are equal and opposite (in 2nd order filters)
)()( jVjV io
C
LCjZL
+Vi(jω)
–
+Vo(jω)
–
ZEQ=∞
C
LCjZC
0
/
||
CL
ZZ
ZZ
ZZZ
CL
CL
CLEQ
Impedances in parallel
ECEN 301 Discussion #20 – Exam 2 Review 78
Basic Filters – Low-Pass Filters
Low-pass Filters: only allow signals under the cutoff frequency (ω0) to pass
)cos()( 1ttvo
0.00. 00 10. 00
.
Low-pass
ω1 ω0 ω2 ω3
)cos(
)cos(
)cos()(
3
2
1
t
t
ttvi
HL(jω)
R
C+
vi(t)–
+vo(t)
–
R
C+
vi(t)–
+vo(t)
–
L
1st Order
2nd Order
ECEN 301 Discussion #20 – Exam 2 Review 79
Basic Filters – High-Pass Filters
High-pass Filters: only allow signals above the cutoff frequency (ω0) to pass
)cos()( 3ttvo )cos(
)cos(
)cos()(
3
2
1
t
t
ttvi
HH(jω)
0.00. 00 10. 00
.
High-pass
ω1 ω0ω2 ω3
+vi(t)
–
+vo(t)
–
1st Order
RC
R
+vi(t)
–
+vo(t)
–
2nd Order
C L
ECEN 301 Discussion #20 – Exam 2 Review 80
Basic Filters – Band-Pass Filters
Band-pass Filters: only allow signals between the passband (ωa to ωb) to pass
)cos()( 2ttvo )cos(
)cos(
)cos()(
3
2
1
t
t
ttvi
HB(jω)
0.00.00 10.00
.
Band-pass
ω1 ωa ω2 ω3ωb
+vi(t)
–
+vo(t)
–
2nd Order
RC
L
ECEN 301 Discussion #20 – Exam 2 Review 81
Basic Filters – Band-Stop Filters
Band-stop Filters: allow signals except those between the stopband (ωa to ωb) to pass
)cos(
)cos()(
3
1
t
ttvo
)cos(
)cos(
)cos()(
3
2
1
t
t
ttvi
HN(jω)
+vi(t)
–
+vo(t)
–
2nd Order
RC
L
0.00. 00 10. 00
.
Band-stop
ω1 ωa ω2 ω3ωb
ECEN 301 Discussion #20 – Exam 2 Review 82
Op-Amps – Open-Loop ModeOpen-Loop Model: an ideal op-amp acts like a difference amplifier (a
device that amplifies the difference between two input voltages)
–
++
v+
–
+
v–
–
+
vo
–
io
i2
i1
–vin
+
–
+
+–
RoutRin
i1
AOLvin
+
vo
–
–
vin
+
v–
v+
NB: op-amps have near-infinite input resistance (Rin) and very small output resistance (Rout)
)(
vvA
vAv
OL
inOLoAOL – open-loop voltage gain
ECEN 301 Discussion #20 – Exam 2 Review 83
Op-Amps – Closed-Loop Mode
Circuit Diagram ACL
Inverting Amplifier
Summing Amplifier
N
nSn
n
F
N
nSnOLno
vR
R
vAv
1
1
–
+ +vo
–
+–
vS
RS
RF
SS
F
SCLo
vRR
vAv
–
+ +vo
–
+–
+–
+–
RSn
RS2
RS1
vSn
vS2
vS1
RF
ECEN 301 Discussion #20 – Exam 2 Review 84
Op-Amps – Closed-Loop Mode
Circuit Diagram ACL
Noninverting Amplifier
Voltage Follower
s
sCLo
v
vAv
SS
F
SCLo
vR
R
vAv
1
–
+ +vo
–
+–
R
RS
RF
vS
–
+ +vo
–
+–
ECEN 301 Discussion #20 – Exam 2 Review 85
Op-Amps – Closed-Loop Mode
Circuit Diagram ACL
Differential Amplifier
12 SSS
Fo vv
R
Rv
–+ +
vo
–
+–
+–
RS
RS
RF
RF
v1
v2
ECEN 301 Discussion #20 – Exam 2 Review 86
Op-Amps – Closed-Loop Mode
Circuit Diagram ACL
Ideal Integrator
Ideal Differentiator
dt
tdvCRv SSFo
)(
t
SFS
o dvCR
v )(1
–
+ +vo(t)
–
+–
vS
CS
RF
–
+ +vo(t)
–
+–
vS
RS
CF
ECEN 301 Discussion #20 – Exam 2 Review 87
Op-Amps
11. find an expression for the gainCF = 1/6 F, R1 = 3Ω, R2 = 2Ω, CS = 1/6 F
+
–
vo(t)iin
CF
R2 v+
v–
iF(t)
i2(t)
R1
i1(t) CS
iS(t)
vs(t)
ECEN 301 Discussion #20 – Exam 2 Review 88
Op-Amps
11. find an expression for the gainCF = 1/6 F, R1 = 3Ω, R2 = 2Ω, CS = 1/6 F
+
–
Vo(jω)Iin
ZF=1/jωCF
Z2 v+
v–
IF(jω)
I2(jω)
Z1
I1(jω)ZS
IS(jω)
Vs(jω)
Node a
Node b
3
1
62
1
2
1
63
1
111111
0
0
:aat KCL
1221
21
21
Soa
SF
oF
a
oa
F
aoaS
F
Vj
Vj
V
ZV
ZZV
ZZZV
Z
VV
Z
VV
Z
VV
III
1. Transfer to frequency domain2. Apply KCL at nodes a and b
NB: v+ = v– and Iin = 0
ECEN 301 Discussion #20 – Exam 2 Review 89
Op-Amps
11. find an expression for the gainCF = 1/6 F, R1 = 3Ω, R2 = 2Ω, CS = 1/6 F
+
–
Vo(jω)Iin
ZF=1/jωCF
Z2 v+
v–
IF(jω)
I2(jω)
Z1
I1(jω)ZS
IS(jω)
Vs(jω)
Node a
Node b
02
1
62
1
0111
00
0
:bat KCL
22
2
2
jVV
ZZV
ZV
Z
V
Z
VV
III
oa
Soa
S
ooa
inS
1. Transfer to frequency domain2. Apply KCL at nodes a and b
ECEN 301 Discussion #20 – Exam 2 Review 90
Op-Amps
11. find an expression for the gain
CF = 1/6 F, R1 = 3Ω, R2 = 2Ω, CS = 1/6 F
033 jVV oa
+
–
Vo(jω)Iin
ZF=1/jωCF
Z2 v+
v–
IF(jω)
I2(jω)
Z1
I1(jω)ZS
IS(jω)
Vs(jω)
1. Transfer to frequency domain2. Apply KCL at nodes a and b3. Express Vo in terms of Vs
Soa VjVjV 235
65
62
j
VV So
ECEN 301 Discussion #20 – Exam 2 Review 91
Op-Amps
11. find an expression for the gain
CF = 1/6 F, R1 = 3Ω, R2 = 2Ω, CS = 1/6 F
+
–
Vo(jω)Iin
ZF=1/jωCF
Z2 v+
v–
IF(jω)
I2(jω)
Z1
I1(jω)ZS
IS(jω)
Vs(jω)
1. Transfer to frequency domain2. Apply KCL at nodes a and b3. Express Vo in terms of VS
4. Find the gain (Vo/VS)
65
62
jV
V
S
o